Saved Bookmarks
| 1. |
A liquid is flowing through horizonatal pipes as shown in figure. |
|
Answer» <P> SOLUTION :The EQUIVALENT circuit can be drawn as under,`X prop (L)/(R^(4)) "" ( "as" X=(8 eta L)/(pi R^(4)))`  `therefore X_(AB) : X_(CD) : X_(EF) : X_(GH)` `=((1)/(2))/(((1)/(2))^(4)):(((1)/(2)))/(((1)/(2))^(4)):((1))/((1)/(2))^(4):((1))/((1)^(4))=8:8:16:1` (a) As the CURRENT is distributed in the inverse ration of the resistance (in parallel). The Q will be distributed in the inverse ratio of X. Thus, volume FLOW through EF will be `(Q)/(17)` and that from GH will be `(16)/(17)Q`. (b) ` X_("net")=8X+[((16X)(X))/((16X)+(X))]+8X=(288)/(17)X` ` therefore Q=(Deltap)/(X_("net")) "" (" as" i=(Delta V)/(R ))` `=((2p_(0)-p_(0)))/((288)/(17)X)=(17p_(0))/(288X)` Now, let `p_(1)` be the pressure at E, then `2p_(0)-p_(1)=8QX=(8xx17 p_(0))/(288)` `therefore "" p_(1)=(2-(17xx8)/(288)) p_(0)=1.53 P_(0)` Similarly, if `p_(2)` be the pressure at F, then `p_(2)-p_(0)=8 QX` `therefore "" p_(2)=p_(0)+(8xx17)/(288)p_(0) " or " p_(2)=1.47 p_(0)` |
|